Logarithm question?

Question

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The answer for x is 3.44, but I do not know how to solve..
Please guide me :(

Answer 1

Note that by the rules of logs, log_a (b) = log b / log a.
With that in mind:

1 / (log_x 2) + 1 / (log_x 3) + 1 / (log_x 6) = 3.6
<=> 1 / (log 2 / log x) + 1 / (log 3 / log x) + 1 / (log 6 / log x) = 3.6
<=> log x / log 2 + log x / log 3 + log x / log 6 = 3.6
<=> log x (1/log 2 + 1/log 3 + 1/log 6) = 3.6
<=> log x (6.703) = 3.6
<=> log x = 0.5371
<=> x = 10^(0.5371) = 3.44.

Answer 2

U have to make base equal
1 / (log_x 2) + 1 / (log_x 3) + 1 / (log_x 6) = 3.6
<=>1 / (log 2 / log x) + 1 / (log 3 / log x) + 1 / (log 6 / log x) = 3.6
<=> log x / log 2 + log x / log 3 + log x / log 6 = 3.6
<=> log x (1/log 2 + 1/log 3 + 1/log 6) = 3.6
<=> log x (6.703) = 3.6
<=> log x = 0.5371
<=> x = 10^(0.5371) = 3.44.

Answer 3

the answer:

1 / (log_x 2) + 1 / (log_x 3) + 1 / (log_x 6) = 3.6
<=>1 / (log 2 / log x) + 1 / (log 3 / log x) + 1 / (log 6 / log x) = 3.6
<=> log x / log 2 + log x / log 3 + log x / log 6 = 3.6
<=> log x (1/log 2 + 1/log 3 + 1/log 6) = 3.6
<=> log x (6.703) = 3.6
<=> log x = 0.5371
<=> x = 10^(0.5371) = 3.44.

Answer 4

Since log a to base b = log a/log b
L.H.S = log x/log 2+log x/log 3+log x/log 6
= log x( 1/log 2+1/log 3+1/log 6)
so: 6.702928578 log x = 3.6
log x = 0.537078675
x = 3.444123175

Answer 5

1 / log(base x) 3 =(log x) / (log 3)

above log can be of any base(>0, /=1)
consider (log x) as variable and solve for it

Answer 6

let log2(basex=a, log3(basex)=b, then log6()basex=a+b
So the given equation becomes:
1/a +1/b +1/(a+b)=3.6or b(a+b)+a(a+b)+ab)=3.6
a^2+b^2+2ab=3.6
(a+b)^2=3.6, So a+b=1.9=log6()basex), or x^1.9=6
1.9 logx = log6=or log x =0.78/1.9=0.41 or x=2.57